如何调试我的"heap corruption"？(C++)

第一次使用，从其他帖子来看，这似乎是一个个案。嗯，我是一个自学成才的C++语言程序员。因此，我终于学会了动态分配内存，并开始思考新的方法来编程计算器，使用CLI输入单个文本字符串并输出答案。这似乎是该项目的一个很好的起点，我开始研究解决方案。以下是我的结果：

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string strInput;
    cin >> strInput;
    int nOperatorCount = 0;
    for(int i = 0; i<(strInput.size());i++)
    {
        if(strInput[i]=='+'||strInput[i]=='-'||strInput[i]=='*'||strInput[i]=='/')
            nOperatorCount++;
    }
    int *pnOperatorLocation = new int[nOperatorCount+1];
    int *pnOperatorType = new int[nOperatorCount];
    float *pnExpressions = new float[nOperatorCount + 1];
    pnOperatorLocation[0] = -1;
    for(int i=0,j=0; i<(strInput.size());i++)
    {
        if(strInput[i]=='+')
    {
            pnOperatorLocation[j+1] = i;
            pnOperatorType[j] = 0;
            pnExpressions[j] = atoi((strInput.substr(pnOperatorLocation[j]+1,i)).c_str());
            j++;
        }
        if(strInput[i]=='-')
        {
            pnOperatorLocation[j+1] = i;
            pnOperatorType[j] = 1;
            pnExpressions[j] = atoi((strInput.substr(pnOperatorLocation[j]+1,i)).c_str());
            j++;
        }
        if(strInput[i]=='*')
        {
            pnOperatorLocation[j+1] = i;
            pnOperatorType[j] = 2;
            pnExpressions[j] = atoi((strInput.substr(pnOperatorLocation[j]+1,i)).c_str());
            j++;
        }
        if(strInput[i]=='/')
        {
            pnOperatorLocation[j+1] = i;
            pnOperatorType[j] = 3;
            pnExpressions[j] = atoi((strInput.substr(pnOperatorLocation[j]+1,i)).c_str());
            j++;
        }
        if(i==(strInput.size()-1))
            pnExpressions[j] = atoi((strInput.substr(pnOperatorLocation[j]+1,i+1)).c_str());
    }
    for(int i=0;i<nOperatorCount+1;i++)
    {
        cout << "pnExpressions[" << i << "]: " << pnExpressions[i] << endl;
    }
    int nOperationsCount = 0;
    for(int i=0;i<nOperatorCount;i++)
    {
        if(pnOperatorType[i]==2||pnOperatorType[i]==3)
        {
            if(pnOperatorType[i+1]==0||pnOperatorType[i+1]==1)
                nOperationsCount++;
            else if((i+1)==nOperatorCount)
                nOperationsCount++;
        }
    }
    cout << "nOperationsCount: " << nOperationsCount << endl;
    float *pnNewExpressions = new float[nOperationsCount];
    for(int i=0,j=0;i<nOperatorCount;i++)
    {
        if(pnOperatorType[i]==2)
        {
            pnNewExpressions[j] = pnExpressions[i] * pnExpressions[i+1];
            if(pnOperatorType[i+1]==2||pnOperatorType[i+1]==3)
            {
                for(int k=i;k<nOperatorCount-i;k++)
                {
                    if(pnOperatorType[k+1]==2)
                        pnNewExpressions[j] = pnNewExpressions[j] * pnExpressions[k+2];
                    else if(pnOperatorType[k+1]==3)
                        pnNewExpressions[j] = pnNewExpressions[j] / pnExpressions[k+2];
                    else
                        break;
                    if(k+1>=nOperatorCount)
                        break;
                }
            }
            j++;
        }
        if(pnOperatorType[i]==3)
        {
            pnNewExpressions[j] = pnExpressions[i] / pnExpressions[i+1];
            if(pnOperatorType[i+1]==2||pnOperatorType[i+1]==3)
            {
                for(int k=i;k<nOperatorCount-i;k++)
                {
                    if(pnOperatorType[k+1]==2)
                        pnNewExpressions[j] = pnNewExpressions[j] * pnExpressions[k+2];
                    else if(pnOperatorType[k+1]==3)
                        pnNewExpressions[j] = pnNewExpressions[j] / pnExpressions[k+2];
                    else
                        break;
                    if(k+1>=nOperatorCount)
                        break;
                }
            }
            j++;
        }
        if(i+1>=nOperatorCount)
            break;
    }
    for(int i=0;i<nOperationsCount;i++)
    {
        cout << "pnNewExpressions[" << i << "]: " << pnNewExpressions[i] << endl;
    }
    float fEvaluation;
    if(pnOperatorType[0]==2||pnOperatorType[0]==3)
        fEvaluation = pnNewExpressions[0];
    else
        fEvaluation = pnExpressions[0];
    cout << "fEvaluation: " << fEvaluation << endl;
    for(int i=0,j=1;i<nOperatorCount;i++)
    {
        if(pnOperatorType[i]==0)
        {
            if(pnOperatorType[i+1]==2||pnOperatorType[i+1]==3)
            {
                fEvaluation = fEvaluation + pnNewExpressions[j];
                j++;
                cout << "fEvaluation: " << fEvaluation << endl;
            }
            else
            {
                fEvaluation = fEvaluation + pnExpressions[i+1];
                cout << "fEvaluation: " << fEvaluation << endl;
            }
        }
        if(pnOperatorType[i]==1)
        {
            if(pnOperatorType[i+1]==2||pnOperatorType[i+1]==3)
            {
                fEvaluation = fEvaluation - pnNewExpressions[j];
                j++;
                cout << "fEvaluation: " << fEvaluation << endl;
            }
            else
            {
                fEvaluation = fEvaluation - pnExpressions[i+1];
                cout << "fEvaluation: " << fEvaluation << endl;
            }
        }
    }
    cout << "fEvaluation: " << fEvaluation << endl;
    delete[] pnOperatorLocation;
    delete[] pnNewExpressions;
    delete[] pnOperatorType;
    delete[] pnExpressions;
    system("pause");
    return 0;
}

如果你通读了所有这些，我非常抱歉，但我一直在写第二本，它被简化了，没有那么复杂，但我想知道为什么这本突然停止了工作。它工作了大约10-15次，我使用了一个输入，如"2*3*4+2*3+2"，并返回"CORRUPTION HEAP DETECTED：在正常块#184之后"。所以我的问题是，有人知道我如何根据错误消息进行调试吗？如果没有，有人能找到为什么我的代码适用于1+1和8*8这样的简单方程，而不适用于我想要的长复杂方程吗？